Binary Search Tree Iterator

Implement the `BSTIterator` class that represents an iterator over the **in-order traversal** of a binary search tree (BST): - `BSTIterator(TreeNode root)` Initializes an object of the `BSTIterator` class. The `root` of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST. - `boolean hasNext()` Returns `true` if there exists a number in the traversal to the right of the pointer, otherwise returns `false`. - `int next()` Moves the pointer to the right, then returns the number at the pointer. Notice that by initializing the pointer to a non-existent smallest number, the first call to `next()` will return the smallest element in the BST. You may assume that `next()` calls will always be valid. That is, there will be at least a next number in the in-order traversal when `next()` is called.   **Example 1:** **Input** ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"] [[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []] **Output** [null, 3, 7, true, 9, true, 15, true, 20, false] **Explanation** BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]); bSTIterator.next(); // return 3 bSTIterator.next(); // return 7 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 9 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 15 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 20 bSTIterator.hasNext(); // return False ```   **Constraints:** - The number of nodes in the tree is in the range `[1, 105]`. - `0 <= Node.val <= 106` - At most `105` calls will be made to `hasNext`, and `next`.   **Follow up:** - Could you implement `next()` and `hasNext()` to run in average `O(1)` time and use `O(h)` memory, where `h` is the height of the tree?